Logarithms: MCQ test, Set 6 | Expresswala Academy

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

The logarithm of $\displaystyle \frac{1}{216}$ to the base $6$ is
A $3$ B $\displaystyle \frac{1}{3}$ C $-3$ D $\displaystyle -\frac{1}{3}$

Answer and solution

Answer: (C) $-3$

$\displaystyle \frac{1}{216} = 6^{-3}$

$\displaystyle \therefore \log_{6} \frac{1}{216} = -3$
If $14^x = 196$ , then $x =$
A $2$ B $-2$ C $\displaystyle \frac{1}{2}$ D $14$

Answer and solution

Answer: (A) $2$

$14^{2} = 196$

$\therefore \log_{14} 196 = 2$
The base of a logarithm cannot be negative because
A $(-5)^2$ is a negative number B $\log_{5} 25$ is negative C $(-5)^2 = 25$ and $5^2 = 25$ would make $\log_{-5} 25 = \log_{5} 25$ D a negative base always gives the value $0$

Answer and solution

Answer: (C) $(-5)^2 = 25$ and $5^2 = 25$ would make $\log_{-5} 25 = \log_{5} 25$

$(-5)^2 = 25 \Rightarrow 2 = \log_{-5} 25$ and $5^2 = 25 \Rightarrow 2 = \log_{5} 25$

So $\log_{-5} 25 = \log_{5} 25$ , which has no meaning.
The logarithmic form of $10^{-4} = 0.0001$ is
A $0.0001 = \log_{10}(-4)$ B $-4 = \log_{0.0001} 10$ C $10 = \log_{-4} 0.0001$ D $-4 = \log_{10} 0.0001$

Answer and solution

Answer: (D) $-4 = \log_{10} 0.0001$

$a^x = N \Leftrightarrow x = \log_{a} N$

The power $-4$ is the logarithm of $0.0001$ to the base $10$ .
The exponential form of $\displaystyle \log_{16} \frac{1}{256} = -2$ is
A $\displaystyle 16 = (\frac{1}{256})^{-2}$ B $\displaystyle \frac{1}{256} = 16 \times (-2)$ C $\displaystyle \frac{1}{256} = (-2)^{16}$ D $\displaystyle \frac{1}{256} = 16^{-2}$

Answer and solution

Answer: (D) $\displaystyle \frac{1}{256} = 16^{-2}$

$\log_{a} N = x \Leftrightarrow N = a^x$

So $\displaystyle \frac{1}{256} = 16^{-2}$ .
$2\log_{19} 19 - \log_{21} 1 =$
A $2$ B $1$ C $0$ D $38$

Answer and solution

Answer: (A) $2$

$\log_{19} 19 = 1,\ \log_{21} 1 = 0$

$= 2(1) - 0 = 2$
$\log_{10} 500 - \log_{10} 5 =$
A $2$ B $\log_{10} 500 \div \log_{10} 5$ C $\log_{10} 495$ D $100$

Answer and solution

Answer: (A) $2$

$\displaystyle = \log_{10} \frac{500}{5} = \log_{10} 100$

$= 2$ since $10^{2} = 100$
$\log_{7} 45 - \log_{7} 9 =$
A $\log_{7} 405$ B $\log_{7} 36$ C $\log_{7} 5$ D $\log_{7} 45 \div \log_{7} 9$

Answer and solution

Answer: (C) $\log_{7} 5$

$\displaystyle \log_{a} m - \log_{a} n = \log_{a} \frac{m}{n}$

$= \log_{7} 5$
If $\log_{y^{4}} x^{2} = k\log_{y} x$ , then $k =$
A $-2$ B $8$ C $2$ D $\displaystyle \frac{1}{2}$

Answer and solution

Answer: (D) $\displaystyle \frac{1}{2}$

$\displaystyle \log_{b^n} a^m = \frac{m}{n} \log_{b} a$

$\displaystyle k = \frac{2}{4} = \frac{1}{2}$
$\displaystyle \log_{\frac{1}{6}} 36 =$
A $\displaystyle -\frac{1}{2}$ B $-2$ C $2$ D $\displaystyle \frac{1}{2}$

Answer and solution

Answer: (B) $-2$

$\displaystyle = \log_{6^{-1}} 6^{2} = \frac{2}{-1}\log_{6} 6$

$= -2$
If $\log_{5} 3 = r$ , then $\log_{3} 5 =$
A $\displaystyle -\frac{1}{r}$ B $\displaystyle \frac{1}{r}$ C $-r$ D $r$

Answer and solution

Answer: (B) $\displaystyle \frac{1}{r}$

$\displaystyle \log_{3} 5 = \frac{1}{\log_{5} 3} = \frac{1}{r}$
$\log_{6} 49 \cdot \log_{7} 6 =$
A $2$ B $4$ C $\displaystyle \frac{1}{2}$ D $-2$

Answer and solution

Answer: (A) $2$

$\log_{b} a \cdot \log_{c} b = \log_{c} a$

$= \log_{7} 49 = 2$
$\displaystyle \frac{\log 49}{\log 7} =$
A $7$ B $\displaystyle \frac{1}{2}$ C $3$ D $2$

Answer and solution

Answer: (D) $2$

$\displaystyle \frac{\log a}{\log b} = \log_{b} a$

$= \log_{7} 49 = 2$
$\displaystyle \frac{10^{\log_{10} 56}}{2^{\log_{2} 8}} =$
A $5$ B $7$ C $64$ D $48$

Answer and solution

Answer: (B) $7$

$a^{\log_{a} N} = N$

$\displaystyle = \frac{56}{8} = 7$
$11^{\log_{3} 4} - 4^{\log_{3} 11} =$
A $0$ B $7$ C $-1$ D $1$

Answer and solution

Answer: (A) $0$

$11^{\log_{3} 4} = 4^{\log_{3} 11}$

So the difference is $0$ .
If $\log_{2} x = \log_{2}(8 - x)$ , then $x =$
A $4$ B $8$ C $-4$ D $5$

Answer and solution

Answer: (A) $4$

$\Rightarrow x = 8 - x$

$\Rightarrow 2x = 8$

$\Rightarrow x = 4$
If $\log_{2}(7 - x) = 3$ , then $x =$
A $1$ B $-15$ C $-1$ D $7$

Answer and solution

Answer: (C) $-1$

$\log_{a} N = x \Rightarrow N = a^x$

$7 - x = 2^{3} = 8$

$\Rightarrow x = -1$
If $\log_{7}(x^2 + x + 1) + \log_{7}(x - 1) = 1$ , then
A $x = \sqrt[3]{7}$ B $x = 2$ C $x = \sqrt[3]{6}$ D $x = 8$

Answer and solution

Answer: (B) $x = 2$

$\log_{7}(x^3 - 1) = 1$

$x^3 = 1 + 7 = 8 \Rightarrow x = 2$
$\log_{3} \log_{2} \log_{2} 256 =$
A $8$ B $1$ C $3$ D $0$

Answer and solution

Answer: (B) $1$

$\log_{2} 256 = 8,\ \log_{2} 8 = 3$

$\log_{3} 3 = 1$
$\displaystyle \frac{\log 3\sqrt[3]{3} + \log 4\sqrt[3]{4}}{\log 12} =$
A $\displaystyle \frac{3}{4}$ B $\displaystyle \frac{1}{3}$ C $1$ D $\displaystyle \frac{4}{3}$

Answer and solution

Answer: (D) $\displaystyle \frac{4}{3}$

$\displaystyle 3\sqrt[3]{3} = 3^{\frac{4}{3}},\ 4\sqrt[3]{4} = 4^{\frac{4}{3}}$

$\displaystyle = \frac{\log 12^{\frac{4}{3}}}{\log 12} = \log_{12} 12^{\frac{4}{3}} = \frac{4}{3}$
$\log_{8} 9 \cdot \log_{27} 4 =$
A $\displaystyle \frac{2}{3}$ B $1$ C $\displaystyle \frac{4}{9}$ D $\displaystyle \frac{9}{4}$

Answer and solution

Answer: (C) $\displaystyle \frac{4}{9}$

$\displaystyle \log_{2^3} 3^2 = \frac{2}{3}\log_{2} 3,\ \log_{3^3} 2^2 = \frac{2}{3}\log_{3} 2$

$\displaystyle = \frac{4}{9}\log_{2} 3 \log_{3} 2 = \frac{4}{9}$
$7^{2 + \log_{7} 2} =$
A $98$ B $343$ C $28$ D $51$

Answer and solution

Answer: (A) $98$

$= 7^2 \times 7^{\log_{7} 2}$

$= 49 \times 2 = 98$
If $\log(2a + b) = \log a - \log b$ , then $a =$
A $\displaystyle \frac{b^2}{2b - 1}$ B $\displaystyle \frac{b}{1 - 2b}$ C $\displaystyle \frac{b^2}{1 - 2b}$ D $\displaystyle \frac{2b^2}{1 - b}$

Answer and solution

Answer: (C) $\displaystyle \frac{b^2}{1 - 2b}$

$\displaystyle 2a + b = \frac{a}{b} \Rightarrow 2ab + b^2 = a$

$\displaystyle \Rightarrow a(1 - 2b) = b^2 \Rightarrow a = \frac{b^2}{1 - 2b}$
If $\displaystyle \log_{\frac{1}{x}} 27 = -3$ , then
A $x = 3$ (or) $x = -3$ B $x = 3$ C $x = -3$ D $\displaystyle x = \frac{1}{3}$

Answer and solution

Answer: (B) $x = 3$

$\displaystyle 27 = \left(\frac{1}{x}\right)^{-3} = x^{3}$

$\Rightarrow x = 3$
On the graph of $\displaystyle y = \log_{\frac{1}{3}} x$ , the value of $y$ at $\displaystyle x = \frac{1}{27}$ is
A $\displaystyle \frac{1}{3}$ B $\displaystyle -\frac{1}{3}$ C $-3$ D $3$

Answer and solution

Answer: (D) $3$

$\displaystyle y = \log_{\frac{1}{3}} \frac{1}{27}$

$= 3$

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MCQ test, Set 6